Verified Answer :

  • To simplify \(j^{14}\), use the cyclic property of the imaginary unit powers, yielding \(j^{14} = j^2 = -1\).
  • To solve \(\frac{10}{5} - \frac{3}{2}\), simplify the first term to 2, then subtract the fraction to get \(\frac{1}{2}\).
  • Final results: \(\boxed{-1}\) and \(\boxed{\frac{1}{2}}\).

Explanation

  1. Analysis of the Expressions

The problem asks for the simplification of two distinct mathematical expressions: \(j^{14}\) (where \(j\) is the imaginary unit satisfying \(j^2 = -1\)) and the arithmetic expression \(\frac{10}{5} - \frac{3}{2}\). We will handle each part individually to provide a clear, step-by-step simplification.

  1. Simplifying \(j^{14}\)

The powers of the imaginary unit \(j\) follow a cyclic pattern:

  • \(j^1 = j\)
  • \(j^2 = -1\)
  • \(j^3 = j^2 \times j = -j\)
  • \(j^4 = j^2 \times j^2 = (-1) \times (-1) = 1\)

Since the pattern repeats every 4 powers, we can simplify \(j^{14}\) by dividing the exponent 14 by 4:
\(14 = 4 \times 3 + 2\).
Thus, \(j^{14} = (j^4)^3 \times j^2\).
Since \(j^4 = 1\), this simplifies to:
\(j^{14} = (1)^3 \times j^2 = 1 \times (-1) = -1\).

  1. Evaluating the Arithmetic Expression

To solve the expression \(\frac{10}{5} - \frac{3}{2}\), we first simplify the fraction \(\frac{10}{5}\):
\(\frac{10}{5} = 2\).
Now the expression is \(2 - \frac{3}{2}\).
To subtract, we find a common denominator, which is 2:
\(2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{4 - 3}{2} = \frac{1}{2}\).
So, the result of the arithmetic expression is \(\frac{1}{2}\) or \(0.5\).

  1. Conclusion

We have successfully simplified both parts of the problem:

  1. The simplification of \(j^{14}\) resulted in \(-1\).
  2. The evaluation of \(\frac{10}{5} - \frac{3}{2}\) resulted in \(\frac{1}{2}\).

Thus, the final results are \(-1\) and \(\frac{1}{2}\).

Examples

The concept of powers of \(j\) is essential in electrical engineering, particularly in analyzing AC circuits where voltages and currents are represented as complex numbers. Simplifying arithmetic fractions is a fundamental skill in everyday budgeting, cooking with measurements, or calculating distances. For example, if a recipe calls for \(2\) cups of flour and you need to subtract a half cup because you are scaling down the recipe, understanding how to subtract fractions ensures you maintain the correct ratios.

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